View source for Expectation of the geometric distribution

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{{ProbMacros}}{{M|\newcommand{\d}[0]{\mathrm{d} } }}
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==Statement==
Let {{M|X\sim}}[[Geometric distribution|{{M|\text{Geo} }}]]{{M|(p)}} where {{M|p}} is the [[probability]] of any trial being a success, and each trial is {{iid}} as {{M|X_i\sim}}[[Borv|{{M|\text{Borv} }}]]{{M|(p)}}, from this we have:
* For {{M|k\in\mathbb{N}_{\ge 1} }} that {{M|\P{X\eq k}\eq p(1-p)^{k-1} }} 
We now define {{M|q:\eq 1-p}} as this will simplify calculations further on, meaning that now:
* For {{M|k\in\mathbb{N}_{\ge 1} }} that {{M|\P{X\eq k}\eq pq^{k-1} }}


'''[[Expectation]]'''
* The expectation of {{M|X}} is: 
** {{MM|\sum^\infty_{k\eq 1}k\P{X\eq k} }} which of course is actually a [[limit]] of a [[series]], {{MM|\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}k\P{X\eq k}\right)}}
: We claim that that {{M|\E{X}\eq\frac{1}{p} }} for {{M|p\in}}[[Half-open-half-closed interval|{{M|(0,1]}}]]{{M|\subseteq\mathbb{R} }} and undefined for {{M|p\eq 0}}


To do so we will consider the 3 cases, {{M|p\eq 0}}, {{M|p\in (0,1)\subseteq\mathbb{R} }} and {{M|p\eq 1}} separately and in reverse of this order.
==See also==
* [[Variance of the geometric distribution]]
* [[Mdm of the geometric distribution]]
==Proof==
We introduce the following for short.
# {{MM|S'_n:\eq\sum^n_{k\eq 1}kpq^{k-1} }} - this forms the [[sequence]] used in the {{link|limit|sequence}} - which is a [[series]]. 
#* Thus {{MM|\E{X}\eq\lim_{n\rightarrow\infty}\Big(S'_n\Big)}}
# {{MM|S_n:\eq\sum^n_{k\eq 1}kq^{k-1} }}
#* This comes from the sequence inside the limit, {{MM|\sum^n_{k\eq 1}k\P{X\eq k}\eq\sum^n_{k\eq 1}kpq^{k-1}\eq p\sum^n_{k\eq 1} kq^{k-1} \eq pS_n}}, so:
#** {{MM|\E{X}\eq\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}k\P{X\eq k}\right)\eq\lim_{n\rightarrow\infty}\Big(pS_n\Big)}}
Notice that {{M|S'_n\eq pS_n}} - introduced purely to save typing.

===Case 1: {{M|p\eq 1}}===
Notice that in this case, {{M|q\eq 1-p\eq 0}}.

We now consider the {{M|S'_n}} terms:
* {{MM|S'_n\eq pS_n\eq p\left(\sum^n_{k\eq 1}kq^{k-1}\right)}} - {{0^0}} comes up here
===Case 2: {{M|p\in (0,1)\subseteq\mathbb{R} }} - {{XXX|EXTENSION}}===
: {{XXX|This case can be extended to {{M|p\in (0,1]}} and should be}} as there's no reason we can't cope with the {{M|q\eq 0}} case - {{XXX|SORT THIS OUT}}
'''[[Lemmas|Lemmas used]]:'''
# {{MM|\frac{\mathrm{d} }{\mathrm{d}q}\Big[q^k\Big]\Bigg\vert_q\eq kq^{k-1} }} which is the first result covered in [[differentiation]]<ref group="Note">I'd really like to link to something here so {{XXX|Link to the actual result!}}</ref>
# {{MM|\sum^n_{k\eq 1}r^{k-1}\eq \frac{1-r^n}{1-r} }} (from the result on the ''[[geometric series]]'' page), and,
#* Note that {{MM|\sum_{k\eq 1}^nr^k\eq r\sum^n_{k\eq 1}r^{k-1} }} so we will really use:
#** {{MM|\sum^n_{k\eq 1}r^k\eq r\frac{1-r^n}{1-r} }}


'''Proof:'''
* Let {{M|p\in}}[[open interval|{{M|(0,1)}}]]{{M|\subseteq}}[[Reals|{{M|\mathbb{R} }}]] be given, and let {{M|X\sim}}[[Geometric distribution|{{M|\text{Geo} }}]]{{M|(p)}} so {{M|\P{X\eq k}:\eq (1-p)^{k-1}p}} for {{M|k\in\mathbb{N}_{\ge 1} }}, now:
** {{MM|\E{X}:\eq\sum^\infty_{k\eq 1}k\cdot\P{X\eq k}\eq \sum^\infty_{k\eq 1}k(1-p)^{k-1}p\eq p\sum^\infty_{k\eq 1}kq^{k-1} }} where we have substituted {{M|q^{k-1} }} for {{M|(1-p)^{k-1} }} at the end there.
*** We use the first lemma described above to observe that {{MM|kq^{k-1}\eq\frac{\d }{\d q}\Big[q^k\Big]\Big\vert_q}}, thus:
**** {{MM|\E{X}\eq p\sum^\infty_{k\eq 1}\frac{\d}{\d q}\Big[q^k\Big]\Big\vert_q}}
****: {{MM|\eq p\cdot\left(\frac{\d}{\d q}\left[\sum^\infty_{k\eq 1}q^k\middle]\right\vert_q\right)}}<ref group="Note" name="CalculusSum"/>
** We now work on the expression: {{MM|\sum^\infty_{k\eq 1}q^k}}, taking it as {{MM|\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}q^k\right) }} and operate on the {{MM|\sum^n_{k\eq 1}q^k}} first
*** By the ''second lemma'' above: 
**** {{MM|\sum^n_{k\eq 1}q^k\eq q\frac{1-q^n}{1-q} }}
****: {{MM|\eq \frac{q}{1-q}\cdot(1-q^n) }}
*** Now we consider {{MM|\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}q^k\right) }},
**** {{MM|\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}q^k\right) \eq \lim_{n\rightarrow\infty}\left(\frac{q}{1-q}\cdot(1-q^n) \right) }}
****: {{MM|\eq\frac{q}{1-q}\cdot\lim_{n\rightarrow\infty}\Big((1-q^n)\Big) }}
**** Let us operate on the {{MM|\lim_{n\rightarrow\infty}\Big((1-q^n)\Big) }} now
***** We have three cases, {{M|q\eq 0}}, {{M|q\in(0,1)}} and {{M|q\eq 1}} - {{Notice|But as we are explicitly in the {{M|q\in(0,1)}} case we don't need to consider them really, we do so for demonstration purposes only}}
****: All of these are applications of [[limit of integer powers of a real value]]
****:# {{M|q\eq 0}} then obviously {{M|0^n}} for {{M|n\in\mathbb{N}_{\ge 1} }} is always {{M|0}} (with [[0^0|{{M|0^0}}]] "disputed" but not relevant here) so
****:#* {{M|\lim_{n\rightarrow\infty}(1-q^n)\eq 1-0\eq 1}}
****:# {{M|q\in(0,1)}} then {{M|q^n}} gets smaller as {{M|n}} increases so {{M|q^n\rightarrow 0}} so {{M|1-q^n\rightarrow 1}}, thus
****:#* {{M|\lim_{n\rightarrow\infty}(1-q^n)\eq 1-0\eq 1}} also
****:# {{M|q\eq 1}} then {{M|q^n\eq 1}} always so
****:#* {{M|\lim_{n\rightarrow\infty}(1-q^n)\eq 1-1\eq 0}}
***** So we see that {{M|q\in [0,1) }}<ref group="Note">I'm extending the range slightly but as {{M|(0,1)\subseteq [0,1)}} we're fine to do so</ref> means that {{M|\lim_{n\rightarrow\infty}(1-q^n)\eq 1}}
**** Substituting our findings we see for the relevant range of this case that:
***** {{MM|\frac{q}{1-q}\cdot\lim_{n\rightarrow\infty}\Big((1-q^n)\Big) \eq \frac{q}{1-q} }}
**** Thus:
***** {{MM|\sum^\infty_{k\eq 1}q^k:\eq \lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}q^k\right) \eq \frac{q}{1-q} }}
** We combine this into our expression for {{M|\E{X} }}:
*** {{MM|\E{X}\eq p\cdot\left(\frac{\d}{\d q}\left[\sum^\infty_{k\eq 1}q^k\middle]\right\vert_q\right)}}
***: {{MM|\eq p\cdot\frac{\d}{\d q}\left[\frac{q}{1-q}\middle]\right\vert_q}}
*** We now operate on {{M|\frac{\d}{\d q}\big[q\cdot (1-q)^{-1}\big]\big\vert_q }} and - as writing it this way implies - will use the [[product rule]]:
**** {{MM|\frac{\d}{\d q}\Big[q\cdot (1-q)^{-1}\Big]\Big\vert_q \eq q\frac{\d}{\d q}\left[(1-q)^{-1}\middle]\right\vert_q+\frac{1}{1-q}\frac{\d}{\d q}\left[q\right]\Big\vert_q}}
****: {{MM|\eq\frac{-q}{(1-q)^2}\cdot\frac{\d}{\d q}\Big[(1-q)\Big]\Big\vert_q +\frac{1}{1-q} }} - notice the [[chain rule|chain ruling]] being applied here
****: {{MM|\eq\frac{-q}{(1-q)^2}(-1) +\frac{1}{1-q} }}
****: {{MM|\eq \frac{1}{1-q}\left(1+\frac{q}{1-q}\right) }}
****: {{MM|\eq \frac{1}{1-q}\left(\frac{1-q}{1-q}+\frac{q}{1-q}\right) }}
****: {{MM|\eq \frac{1}{1-q}\left(\frac{1}{1-q}\right) }}
****: {{MM|\eq\frac{1}{(1-q)^2} }} or {{M|\eq (1-q)^{-2} }}
**** Finally: {{MM|\frac{\d}{\d q}\big[q\cdot (1-q)^{-1}\big]\big\vert_q \eq \frac{1}{(1-q)^2} }}
***** So now we have: {{MM|\E{X} \eq p\cdot\frac{\d}{\d q}\left[\frac{q}{1-q}\middle]\right\vert_q\eq p\frac{1}{(1-q)^2} }}
*** Lastly we operate on {{MM|\E{X}\eq p\frac{1}{(1-q)^2} }}
**** Recall that {{M|q:\eq 1-p}} so {{M|1-q\eq 1-(1-p)\eq 1-1+p\eq p}} - we have {{M|1-q\eq p}} now, substitute this in and we see:
***** {{MM|\E{X}\eq p\frac{1}{p^2} }}
*****: {{MM|\eq \frac{1}{p} }}
** So we have {{MM|\E{X}\eq \frac{1}{p} }} given our value of {{M|p}}
* Since our choice of {{M|p\in(0,1)}} was arbitrary we have shown:
** {{MM|\forall p\in(0,1)\left[\E{\text{Geo}(p)}\eq\frac{1}{p}\right] }} - as required
==Notes==
<references group="Note">
<ref group="Note" name="CalculusSum">Remember {{MM|\sum^\infty_{k\eq 1}a_k}} is just short hand for {{MM|\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}a_k\right)}} - see [[limits]] and [[limit of a series]] - and remember that {{MM|\frac{\d}{\d x}\Big[f(x)\Big]\Big\vert_x+\frac{\d}{\d x}\Big[g(x)\Big]\Big\vert_x\eq\frac{\d}{\d x}\Big[f(x)+g(x)\Big]\Big\vert_x}} - as per [[linearity of the derivative]].
* {{XXX|More links?}}</ref>
</references>
{{Theorem Of|Probability|Elementary Probability|Statistics}}[[Category:Expectation Calculations]]